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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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A man reaches his office 30 min late, if he walks from his home at 3 km per hour and reaches 40 min early if he walks 4 km per hour. How far is his office from his house?

A3km

B14km

C5km

D7km

Explanation:

Lets assume distance between office and his house =x km
Time taken when speed is 3 kmph = x/3 min.
Time taken when speed is 4 kmph = x/4 min.
Difference between time taken 40-(-30) = 70mins = 70/60 hours.
x/3- x/4 = 70/60
x/12= 7/6
x=14 km

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City bus runs two bus from A to B each making 5 round trips in a day. There are no stops between. These buses ply back and forth on same route at diff. but uniform speed. Each bus start at 7 am from respective terminus. They meet for the first time at 7 km from A. Their next meeting is at distance of 4 km from B, while travelling in opposite direction. Assuming that time by buses at terminus is negligibly small, and cost of running a bus is 20 per km. Find daily cost of buses ?

ARs. 3000

BRs. 6800

CRs. 3400

DRs. 3600

Explanation:

Let's assume total distance travelled by both bus = x km
So as per question

First meet at 7 km from A.
By that time first bus travels 7 Km and second one travels x-7 Km. As they have travelled for the same time,

Second meeting is at 4 km from B:
By that time first bus travels (x+4) Km and second bus will travels x+(x-4) = (2x-4) km.
=> 7/(x+4) = (x-7)/(2x-4)
x^2 - 15x = 0 => x=17 km.
So one round trip = 2*17 = 34 km.
as given each bus make 5 round trips in a day.

so total distance cover by both buses = 34 * 2 *5 = 340 km.
Cost of travelling per bus = 340 * 20 = Rs. 6800.

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A man sets out to cycle from Delhi to Rohtak and at the same time another man starts from Rohtak to cycle to cycle to Delhi. After passing each other they completed their journey in (10/3) hours and (16/3) hours respectively.At what rate does the second man cycle if the first cycle at 8 kmph?

A6.12 kmph

B6.42 kmph

C6.22 kmph

D6.32 kmph

Explanation:

A's Speed:B's Speed=rootof(b):rootof(a)
=>8:x=rootof(10/3):rootof(16/3)
On solving we get,
=> x=2*rootof(10)
=> x=2*3.16
=> x=6.32 kmph.

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The distance between 2 persons is 800m. If they start moving towards each other simultaneously at 10m/s and 15m/s, in how much time do they meet?

A25 sec

B30 sec

C20 sec

D32 sec

Explanation:

Total speed of both person = 10 + 15 = 25 m/s

Distance between both persons = 800m
so they meet after 800/25 = 32 sec.

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A scooterist covers a certain distance at 36 kmph. How many meters does he cover in 2 min?

A800 meters.

B1100 meters.

C700 meters.

D1200 meters.

Explanation:

Speed = 36 kmph = 36 * 5/18 = 10mps
So, Distance covered in 2 min = (10 * 2 * 60)m = 1200 meters.

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The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

A250

B150

C300

D200

Explanation:

Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.

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By walking at 3/4th of his usual speed, a man reaches office 20 minutes later than usual. What is his usual time?

A60 minutes.

B50 minutes.

C30 minutes.

DNone of these

Explanation:

3/4 of a man's usual speed means, he takes 4/3 of his usual time to cover the same distance,
i.e. he takes 4/3 â€“ 1 = 1/3 time extra.
Given 1/3 time is 20 minutes
Usual time = 20 * 3 = 60 minutes.

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If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.

A10 km

B8 km

C6 km

D5 km

Explanation:

Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km

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Yana and Gupta leave points x and y towards y and x respectively simultaneously and travel in the same route. After meeting each other on the way, Yana takes 4 hours to reach her destination, while Gupta takes 9 hours to reach his destination. If the speed of Yana is 48 km/hr, what is the speed of Gupta?

A30 mph

B50 mph

C35 mph

D20 mph

Explanation:

Let the speed of Yana be v1 and Gupta be v2 and time taken by them be t1 and t2 respectively.
There is a shortcut for such questions as v1^2/v2^2= t2/t1

Putting the respective values we get v2= 32km/hr

1.6 km =1 miles

So 32/1.6= 20mph.

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Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

A55 min.

B45 min.

C35 min.

D50 min.

Explanation:

New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) - usual time = 10min
Therefore Usual time = 50min

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